Calculate the flow through a plane

1. Dec 25, 2013

brkomir

1. The problem statement, all variables and given/known data
Let $\vec{F}(x,y,z)$ be defined as $\vec{F}(x,y,z)=(-x+y^2+z^3,xe^y,-xze^y+z+1)$ and plane $\Sigma$ defined with parameterization $\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3u+v)+u,v+(1-u^2-v^2)sinhu,-u^2-v^2+1)$ where u,v are elements of a circle in point (0,0) with a radius 1. Orientation of $\Sigma$ is regulated in accordance with parametrization.

Calculate the flow of $\vec{F}$ through plane $\Sigma$.
2. Relevant equations

3. The attempt at a solution
Hi there!

I have a problem... I have no idea how to start on this problem? I know that I can replace the x,y,z in $\vec{F}$ with u and v from parameterization $\vec{r}$ but this would give me some massive never ending calculations.

Any other way to do this?

2. Dec 25, 2013

Staff: Mentor

Can one think of a relevant equation that describes a flow based on the flux through an area?

What is $\vec{F}$?

Is $\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u+v))+u,v+(1-u^2-v^2) sinh u,-u^2-v^2+1)$ correct?

3. Dec 25, 2013

brkomir

You're probably talking about Gaussian law... but, to be honest with you, I don't know when am I allowed to use it...

Am... there is one little mistake (sorry for that).. it's $\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u)+v)+u,v+(1-u^2-v^2) sinh (u),-u^2-v^2+1)$

4. Dec 25, 2013

HallsofIvy

Staff Emeritus
First, is that really a plane and not a curved surface? I'm not saying it isn't because I haven't checked, but a plane can be written in terms of linear[/b ] functions of the parameters but doesn't have to be . In any case, if it is a plane then the normal vector has constant direction.

The flow of $\vec{F}(x,y,z)= f(x, y, z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ through a surface is the integral of the dot product of $\vec{F}$ and the unit normal vector to the surface at each point.

The "normal vector" to a surface given by $\vec{r}(u,v)$ is given by the cross product of [itex]\vec{r_u}\times \vec{r_v}. The unit normal is, of course, that normal vector divided by its own length.

5. Dec 25, 2013

brkomir

Aaam, I'm not sure if that is a plane or a curved surface. If I would have to guess i'd say it's a curved surface, however i am not really even sure how one can check that?

Ok, but I assume I have to change $\vec{F}(x,y,z)$ into $\vec{F}(u,v)$ or not?

6. Dec 26, 2013

brkomir

$\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u)+v)+u,v+(1-u^2-v^2) sinh (u),-u^2-v^2+1)$

Than $\vec{r_u}=(-2u(u^2+v^2-1)(sinh^3(u)+v)-2u(u^2+v^2)(sinh^3(u))+v)-3(u^2+v^2-1)(u^2+v^2)sinh^2(u)cosh(u)+1,-(u^2+v^2-1)cosh(u)-2usinh(u),-2u)$

and

$\vec{r_v}=(-u^4+u^2(1-6v^2)-2v(2u^2+2v^2-1)sinh^3(u)-5v^4+3v^2,1-2vsinh(u),-2v)$

is there anything I can do before calculating the normal vector $\vec{r_u}\times \vec{r_v}$ ?? Please say yes :D